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1 month ago

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A (1.25+A) kg bowling ball is hung on a (2.50+B) m long rope. It is then pulled back until the rope makes an angle of (12.0+C)o

with the vertical and released. Find the tension in the rope when the ball is at the lowest point. Give your answer in N and with 3 significant figures.

1 answer:

Ostrovityanka [3.2K]1 month ago

7 0

Answer:

F = 0.535 N

Explanation:

We will apply energy concepts, considering both the peak and the bottom of the path.

Top

Em₀ = U = mg y

Bottom

$Em_{f}$ = K = ½ m v²

Emo = $Em_{f}$

mg y = ½ m v²

v = √ (2gy)

y = L - L cos θ

v = √ (2g L (1 - cos θ))

Next, we will employ Newton's second law at the lowest point where the acceleration is centripetal.

F = ma

a = v² / r

For the turning radius, the cable length is r = L.

F = m 2g (1 - cos θ)

Now, let's find the result.

F = 2 1.25 9.8 (1 - cos 12)

F = 0.535 N

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A diver explores a shallow reef off the coast of Belize. She initially swims d1 = 74.8 m north, makes a turn to the east and con

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Response:R=1607556m

θ=180degrees

Clarification:

d1=74.8m

d2=160.7km=160.7km*1000

d2=160700m

d3=80m

d4=198.1m

Utilizing an analytical approach:

Rx=-(160700+75*cos(41.8))= -160755.9m

Ry= -(74.8+75sin(41.8))-198.1=73m

Magnitude, R:

R=√Rx+Ry

R=√160755.9^2+20^2=160755.916

R=160756m

Direction,θ:

θ=arctan(Rx/Ry)

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θ=-7.9256*10^-6

It is worth noting that since θ is in the second quadrant, 180 is added

θ=180-7.9256*10^6=180degrees

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1 month ago

During a snowball fight two balls with masses of 0.4 and 0.6 kg, respectively, are thrown in such a manner that they meet head-o

Yuliya22 [3333]

Answer:

The snowball's speed after the impact is 3 m/s

Explanation:

Given the following:

mass of each ball

m₁ = 0.4 Kg

m₂ = 0.6 Kg

initial speed of both balls = v₁ = 15 m/s

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Applying conservation of momentum

m₁ v₁ - m₂ v₁ = (m₁+m₂) V

A negative velocity indicates that the second ball moves in the opposite direction.

0.4 x 15 - 0.6 x 15 = (1) V

Therefore,

V = - 3 m/s

Consequently,

The snowball's speed following the collision is 3 m/s

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2 months ago

A small cork with an excess charge of +6.0µC is placed 0.12 m from another cork, which carries a charge of -4.3µC.

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A) 16.1 N

The force of electricity acting between the corks can be calculated using Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where

k represents Coulomb's constant

$q_1 = 6.0 \mu C=6.0 \cdot 10^{-6} C$ denotes the charge magnitude on the first cork

$q_2 = 4.3 \mu C = 4.3 \cdot 10^{-6}C$ indicates the charge magnitude on the second cork

r = 0.12 m is the distance separating the corks

By inserting the values into the formula, we arrive at

$F=(9\cdot 10^9 N m^2 C^{-2} )\frac{(6.0\cdot 10^{-6}C)(4.3\cdot 10^{-6} C)}{(0.12 m)^2}=16.1 N$

B) Attractive

<pas per="" coulomb="" law="" the="" orientation="" of="" electric="" force="" between="" two="" charged="" entities="" relies="" on="" their="" charge="" signs.=""><pmore specifically="">

- when both are similarly charged (e.g. positive-positive or negative-negative), the force is repulsive

- when charges are of opposite signs (e.g. positive-negative), the resulting force is attractive

<pin this="" case="" we="" have="">

Cork 1 holds a positive charge

Cork 2 possesses a negative charge

<pthus the="" force="" acting="" between="" them="" is="" attractive.="">

C) $2.69\cdot 10^{13}$

The total charge of the negative cork is

$q_2 = -4.3 \cdot 10^{-6}C$

<pwe understand="" that="" a="" single="" electron="" has="" charge="" of="">

$e=-1.6\cdot 10^{-19}C$

<pthe total="" charge="" of="" the="" negative="" cork="" arises="" from="" having="" n="" extra="" electrons="" so="" we="" can="" express="" it="" as="">

$q_2 = Ne$

<pafter solving="" for="" n="" we="" can="" determine="" the="" count="" of="" excess="" electrons:="">

$N=\frac{q_2}{e}=\frac{-4.3\cdot 10^{-6} C}{-1.6\cdot 10^{-19} C}=2.69\cdot 10^{13}$

D) $3.75\cdot 10^{13}$

The overall charge on the positive cork is

$q_1 = +6.0\cdot 10^{-6}C$

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$q_1 = -Ne$

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$N=-\frac{q_1}{e}=-\frac{+6.0\cdot 10^{-6} C}{-1.6\cdot 10^{-19} C}=3.75\cdot 10^{13}$

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2 months ago

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kicyunya [3294]

To address this question, we will utilize concepts linked to centripetal force, aligning it with the static frictional force acting on the object. Using this relationship, we can derive the velocity and input the known values. The defined values are:

$r = 16m$

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$\mu_s = 0.63$

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$F_c = \frac{mv^2}{r}$

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3 0

2 months ago

A pole-vaulter is nearly motionless as he clears the bar, set 4.2 m above the ground. he then falls onto a thick pad. the top of

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Refer to the diagram below.

Ignoring air resistance, use gravitational acceleration g = 9.8 m/s².

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2 months ago

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